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Difference-Amplifier

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dark

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Hi,

I see a circuit of difference amplifier where the reference is set to 10V (the rest of the circuit require a bipolar supply , therefore the reference) . I know the transfer function of the difference amplifier when referenced to GND , but what is the ckt behaviour in the attached image . When we reference it to 10V , is it the same as before Vout=((V2-V1)(Rf/Rin) , hope V2,V1,Rf,Rin are understood . And this time the ckt is expected to get a VCC of 14V? , is this correct.


Thanks
-Adi
 

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Hi,

I see a circuit of difference amplifier where the reference is set to 10V (the rest of the circuit require a bipolar supply , therefore the reference) . I know the transfer function of the difference amplifier when referenced to GND , but what is the ckt behaviour in the attached image . When we reference it to 10V , is it the same as before Vout=((V2-V1)(Rf/Rin) , hope V2,V1,Rf,Rin are understood . And this time the ckt is expected to get a VCC of 14V? , is this correct.


Thanks
-Adi

hi,
How do you get a value of 14Vout.?
 
hi,
How do you get a value of 14Vout.?

Hi, I was assuming if GND=10V then VCC= 24-10 ? not sure! or is it mid way in between 24V , is it in between 0V----10V----24V.

Regards/
 
When the AC input signal is less than +10V then the output of the opamp will be less than +10V.
When the AC input signal is exactly +10V then the output of the opamp is exactly +10V.
When the AC input signal is more than +10V then the output of the opamp is more than +10V.

R3 and R1 attenuate the input to 0.0909 times.
The gain of the opamp is (150k/10k) + 1= 16. The total gain from input to output is
0.0909 x 16= 1.45 times.
 
When the AC input signal is less than +10V then the output of the opamp will be less than +10V.
When the AC input signal is exactly +10V then the output of the opamp is exactly +10V.
When the AC input signal is more than +10V then the output of the opamp is more than +10V.

R3 and R1 attenuate the input to 0.0909 times.
The gain of the opamp is (150k/10k) + 1= 16. The total gain from input to output is
0.0909 x 16= 1.45 times.

Hi audioguru ,Thanks ;

How do you get ' 0.0909 times.' figure?

It seems to be reasonable what you explain the 10V becomes the center point of the either side swing .

With (V2-v1)*Gain I end up with a -ve sign , what transfer function holdes now.

Regards
-Adi
 
Hi audioguru ,Thanks ;

How do you get ' 0.0909 times.' figure?

It seems to be reasonable what you explain the 10V becomes the center point of the either side swing .

With (V2-v1)*Gain I end up with a -ve sign , what transfer function holdes now.

Regards
-Adi

hi,
Ratio of, 5K/[50K + 5K] = 0.0909
 
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